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I want to prove that $\nabla\times(\nabla\times \boldsymbol{A}) = \nabla(\nabla\cdot\boldsymbol{A}) - \nabla^2\boldsymbol{A}$ using the Levi Civita. This is solved considering the $i$-th component of $\nabla\times(\nabla\times \boldsymbol{A})$. But I was performing the calculation and don't know where my mistake is:

\begin{align} \nabla\times(\nabla\times \boldsymbol{A}) &= \nabla \times (\epsilon_{ijk}\hat{e}_i\nabla_jA_k) =\epsilon_{lmn}\hat{e}_l\nabla_m(\epsilon_{ijk}\hat{e}_i\nabla_jA_k)_n\\ &=\epsilon_{lmn}\hat{e}_l\nabla_m\epsilon_{njk}\nabla_jA_k = \epsilon_{lmn}\epsilon_{njk}\hat{e}_l\nabla_m\nabla_jA_k\\ &=(\delta_{lk}\delta_{mj} - \delta_{km}\delta_{lj})\hat{e}_l\nabla_m\nabla_jA_k\\ &=\hat{e}_l\nabla_m\nabla_mA_l - \hat{e}_l\nabla_m\nabla_lA_m = \nabla^2\boldsymbol{A} - \nabla(\nabla\cdot\boldsymbol{A}). \end{align}

I get the result but with a negative sign. The only part in which I may have done things badly is when taking the levi-civita product, but I'm pretty sure that I did that correctly.

Does somebody spot my mistake? I appreciate your effort.

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  • $\begingroup$ The bracket with the deltas appears to be the wrong way round, and this is where the errant minus sign is coming from $\endgroup$ – David Quinn Feb 21 '16 at 23:16
  • $\begingroup$ @DavidQuinn but if you compute $\epsilon_{lmn}\epsilon_{njk}$ with the determinant you get the expression of deltas I wrote. $\endgroup$ – Vladimir Vargas Feb 21 '16 at 23:21
  • $\begingroup$ So shouldn't that be $\delta_{lj}\delta_{mk}-\delta_{lk}\delta_{mj}$? $\endgroup$ – David Quinn Feb 21 '16 at 23:29
  • $\begingroup$ @DavidQuinn look, here there is a picture of how I computed the determinant: postimg.org/image/sswnt7qmv $\endgroup$ – Vladimir Vargas Feb 21 '16 at 23:40
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    $\begingroup$ The mistake is indeed in computing the determinant. Try to expand the determinant first and then contract the $n$ index in the end (you seem to be doing both at once). For example you seem to have taken $\delta_{nn} = 1$ but this term should be $3 = \sum_{n=1}^3\delta_{nn}$. $\endgroup$ – Winther Feb 25 '16 at 14:07
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In order to leave this question answered I will point out that @Winther's answer is correct. I took $\delta_{nn}=1$, but this term should be $3$, following the Einstein notation for summation.

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